Problem: Let $f$ be a transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} 0 & -z\sin(y) & \cos(y) \\ \\ \cos(z) & 0 & -x\sin(z) \\ \\ -y\sin(x) & \cos(x) & 0 \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $\left( \dfrac{\pi}{3}, 0, \dfrac{\pi}{3} \right)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely
Solution: The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} 0 & -z\sin(y) & \cos(y) \\ \\ \cos(z) & 0 & -x\sin(z) \\ \\ -y\sin(x) & \cos(x) & 0 \end{bmatrix} \right) \\ \\ &= z\sin(y)(0-xy\sin(z)\sin(x)) + \cos(y)(\cos(z)\cos(x)-0) \\ \\ &= -xyz\sin(x)\sin(y)\sin(z) + \cos(x)\cos(y)\cos(z) \end{aligned}$ If we evaluate $|J(f)|$ at $\left( \dfrac{\pi}{3}, 0, \dfrac{\pi}{3} \right)$, we get $\dfrac{1}{4}$. Because the Jacobian determinant here has an absolute value less than $1$ but not $0$, we can conclude that $f$ will finitely contract the space around $\left( \dfrac{\pi}{3}, 0, \dfrac{\pi}{3} \right)$. To recap, the Jacobian determinant of $f$ is $-xyz\sin(x)\sin(y)\sin(z) + \cos(x)\cos(y)\cos(z)$ and $f$ will finitely contract the space around the point $\left( \dfrac{\pi}{3}, 0, \dfrac{\pi}{3} \right)$.